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Tcecarenko [31]
2 years ago
8

A rectangular field has an area of 1,764 m2. The width of the field is 13 m more than the length. What is the perimeter of the f

ield?

Mathematics
2 answers:
leva [86]2 years ago
6 0

Answer:

170.

Step-by-step explanation:

We have that A = W*L where A= Area, W= width, L=length and that W = 13+L. So,

A = (13+L)*L

1764 = 13L +  L^2

L^2 + 13L - 1764 = 0.

We use the general formula in the image where a=1, b=13 and c= -1764.

L1= \frac{-13+\sqrt[2]{169-4*(-1764)} }{2}

L1= \frac{-13+\sqrt[2]{7225} }{2}

L1= \frac{-13+85}{2}

L1= 36.

L2= \frac{-13-\sqrt[2]{169-4*(-1764)} }{2}

L2= \frac{-13-85}{2}, this result will be negative and the problem is about length so it doesn't apply for this problem. We use L1= 36.

Then, L=36, W = 13 + 36 = 49. Therefore, the perimeter is 2L+2W = 72+98=170.

Gnoma [55]2 years ago
4 0
Hello,
Here in France we call length the greater side of a rectangle.
So:
Let a the greater side
b the shortest

a=b+13
a*b=1764
==>b*(b+13)=1764
==>b²+13b-1764=0
==>b=(-13-85)/2 to exclude
or (b=(-13+85)/2=36 and a=36+13=49)

P=2(a+b)=2(36+49)=170 (m)
Proof:
49-36=13
36*49=1764


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